# Thread: Geometry figure, sum of distances to three points

1. ## Geometry figure, sum of distances to three points

Looking to definition of an ellipse we can see that it is the set of points which the sum of distances to two fixed points is a constant.
So i was thinking about a geometry figure that is the set of points which the sum of distances to three fixed points is a constant. Does this geometry figure exist?
To check i tried to find a equation to this figure.

Taking the points A,B,C creating an equilateral triangle. With A and C in the X axis and B in the Y axis. The side of the triangle is (2*a).
So A(-a,0), C(a,0) and B(0,a*sqrt(3) )
I used these definitions to simplify the work.
Being P(x,y) an generic point, and 2a the constant distance arbitrary choosed.

(Distance PA) + (Distance PB) + (Distance PC) = 2a

sqrt((x+a)²+(y)²) + sqrt((x)²+(y-a*sqrt(3))²) + sqrt((x-a)²+y²) = 2a

Developing the polinomial I got this tiny thing:
x²+y²-a² +2a(2x+2*sqrt(x²-2ax+a²+y²)-sqrt(3)*y) +2*sqrt(x^4+2x²y²+2ax³+4a²x²-2a*sqrt(3)*yx²+2axy²-4a²*sqrt(3)*xy
+6a³x-2a³*sqrt(3)*y+4a²y²+3a^4+y^4-2a*sqrt(3)*y^3)) = 0

Putting this equation in a grapher like aGrapher, i got nothing. So i would like to know if this geometry figure i idealize exists or not, and if it exists how does it look like and why my equation doesnt work.

Thank you so much, forgive my mistakes in grammar im not native.

2. Hello, Boyrog!

I tried it myself and got into an awful mess . . .

So i was thinking about a geometry figure that is the set of points
which the sum of distances to three fixed points is a constant.
Does this geometry figure exist?
I didn't bother with an equilateral triangle.
I used simpler coordinates.
Code:
               C|
o (0,a)
* | *
*   |   *
*     |     *
B  *       |       *  A
- - o - - - - + - - - - o - -
(-a,0)       |       (a,0)

We want point $\displaystyle P(x,y)$ so that: .$\displaystyle \overline{PA} + \overline{PB} + \overline{PC} \:=\:k$

So: .$\displaystyle \sqrt{(x-a)^2+y^2} + \sqrt{(x+a)^2+y^2} + \sqrt{x^2 + (y-a)^2} \:=\:k$

I see no practical way to eliminate the radicals.

Maybe you want to square the equation repeatedly.
. . I'll wait in the car . . .
.

3. Thank you for helping Soroban,
your definition has helped me. I checked the equation and corrected some mistakes I had done. The correct devoloping is:

PA + PB + PC = K
Code:
((x-a)^2+y^2)^0.5+((x+a)^2+y^2)^0.5+(x^2+(y-a)^2)^0.5= K

Isolating the roots and elevating to square till get no square roots:
Code:
[(k²-x²-a²-y²-2ay)²-4((x²-a²)²+y²(2x²+2a²)+y^4)-4k²(x²+(y-a)²)]² - 64k²[(x²-a²)²+y²(2x²+2a²)+y^4)]*[x²+(y-a)²]
Putting

this equation in the grapher, we get this kind of flower: (very nice isn't? )